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Prove that : $\dfrac{n+1}{2} \leq 2\cdot\sqrt{2}\cdot\sqrt[3]{2}\cdot\sqrt[4]{2}\cdots\sqrt[n]{2}$.

I am unable to prove this even by induction and general method. Indeed, when I look at the question $2\cdot\sqrt{2}\cdot\sqrt[3]{2}\cdot\sqrt[4]{2}\cdots\sqrt[n]{2}\leq n+1$, asked by me, I have received a hint as a comment to use binomial theorem and showed $$\left(1+\frac1n\right)^n=\sum_{k=0}^n{n\choose k}\frac1{n^k}\geq1+{n\choose 1}\frac1n=2.$$ So, expression becomes $$2 \cdot \sqrt{2} \cdot \sqrt[3]{2} \cdots \sqrt[n]{2} \leq \left(1 +\dfrac{1}{1}\right)\left(1 +\dfrac{1}{2}\right)\left(1 +\dfrac{1}{3}\right)...\left(1 +\dfrac{1}{n}\right)=n+1.$$ I want to solve this problem exactly by same method. So, for this I've to prove that $$\sqrt[n]{2} \leq \dfrac{n+2}{n+1}.$$ How do I do this ?

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    $\begingroup$ The inequality simply isn't true. For large $n$ the RHS will be $O(2^{\ln n}) = O(n^c)$ for some $c<1$. $\endgroup$
    – Erick Wong
    Commented Nov 13, 2015 at 7:22
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    $\begingroup$ It's false with $n=34$ (and above). $\endgroup$
    – 2'5 9'2
    Commented Nov 13, 2015 at 7:36
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    $\begingroup$ What main result? And @EwanDelanoy, what argument? I'm just saying it's false, agreeing with Erick Wong. $\endgroup$
    – 2'5 9'2
    Commented Nov 13, 2015 at 7:40
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    $\begingroup$ @EwanDelanoy Oh I see. I deleted that comment like 5 seconds after posting it. Erick's $c$ is actually $1/\log_2(e)$, not $\log_2(c)$ as I mistakenly thought for a second. $\endgroup$
    – 2'5 9'2
    Commented Nov 13, 2015 at 7:48
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    $\begingroup$ Oh, I'm not angry at all. I just don't understand what you are after. If you raise $2$ to the left and center parts of this inequality, you get $(k+1)^{1/2}\leq2\cdot\sqrt{2}\cdots\sqrt[k]{2}$. So you are mistaking what the left side becomes. It's not $\frac{k+1}{2}$. $\endgroup$
    – 2'5 9'2
    Commented Nov 13, 2015 at 8:02

1 Answer 1

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Considering the rhs, $$A_n=\prod_{i=1}^n 2^{\frac 1 n}$$ $$\log(A_n)=\sum_{i=1}^n \log(2^{\frac 1 n})=\log(2)\sum_{i=1}^n \frac 1 n=H_n\log(2)$$ For large values of $n$, $$H_n=\gamma +\log(n)+\frac{1}{2 n}+O\left(\frac{1}{n^2}\right)$$ while the logarithm of the lhs would write $$\log(2)+\log(n) +\frac{1}{n}+O\left(\frac{1}{n^2}\right)$$ and, using these short expansions, you could see that $\left(\log(rhs)-\log(lhs)\right)$ cancels close to $n=32$ and from that point, becomes negative.

So, as comments already showed it, the inequality only holds for a small range of $n$ (up to some finite $n$ as Did properly pointed out).

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    $\begingroup$ Asymptotic expansions cannot provide information about the rank $n$ at which RHS becomes smaller than LHS. Replacing "up to $n=33$" by "up to some finite $n$", the approach works. The alternative is to use nonasymptotic estimates of the harmonic numbers and of the logarithm (which is not difficult in the present case). $\endgroup$
    – Did
    Commented Nov 13, 2015 at 9:01
  • $\begingroup$ @Did. I totally agree and I shall edit my answer accordingly. $\endgroup$
    – Claude Leibovici
    Commented Nov 13, 2015 at 9:03
  • $\begingroup$ I don't understand... the lhs is simply $\frac{n+1}{2}$, so isn't its logarithm $\log(n+1) - \log(2)$? $\endgroup$
    – A.P.
    Commented Nov 13, 2015 at 9:12
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    $\begingroup$ @A.P. In order to have similar terms I wrote $\log(n+1)=\log(n)+\log(1+\frac 1n)$ and expanded the second logarithm. $\endgroup$
    – Claude Leibovici
    Commented Nov 13, 2015 at 9:22

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